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3 years ago

While traveling along a highway a driver slows from 24 m/sec to 15 m/sec in 12 seconds. What is the

Physics

1 answer:

Vladimir [108]3 years ago

4 0

24-15=9 m/s slower in 12 seconds. So 9/12 m/s² slower. Therefore the acceleration is -0,75 m/s²

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How much work is needed to lift a 3kg create a vertical displacement of 22m

Ulleksa [173]

I think its W= 647.239J

6 0

3 years ago

A sound source is located somewhere along the x-axis. Experiments show that the same wave front simultaneously reaches listeners

galina1969 [7]

Answer:

Explanation:

As the source is situated on x - axis , it must be situated in between the two listeners .

So the x coordinate of source is

(-7 + 3 )/2

= - 2 m

The equation of the wave- front will be that o a circle having centre at (-2,0)

and radius = distance between -2 and 3 , that is 5 m

equation of circle

=( x+2 )² + y² = 25

It cuts y axis when x = 0

Putting x = 0

4 + y² = 25

y² = 21

y = + √21 , or - √21

7 0

3 years ago

The force on the spring is F0 and it stores elastic potential energy PEs0. If the spring displacement is tripled to 3x0, determi

Dimas [21]

Answer:

Explanation:

Let initial extension in the spring= x₀

Force on the spring = F₀

Let spring constant = k

Fo = k x₀

Fn = 3k x₀

Fn /Fo = 3

PEs0 ( ORIGINAL) =1/2 k x₀²

PEsn ( NEW) =1/2 k (3x₀)²

PEsn / PEs0 = 9

7 0

3 years ago

The _______ is the time required for one complete wave oscillation to occur.

aleksandrvk [35]

Period is the answer

6 0

3 years ago

Read 2 more answers

A 0.20-kg mass is oscillating on a spring over a horizontal frictionless surface. When it is at a displacement of 2.6 cm for equ

valentinak56 [21]

Explanation:

The given data is as follows.

mass = 0.20 kg

displacement = 2.6 cm

Kinetic energy = 1.4 J

Spring potential energy = 2.2 J

Now, we will calculate the total energy present present as follows.

Total energy = Kinetic energy + spring potential energy

= 1.4 J + 2.2 J

= 3.6 Joules

As maximum kinetic energy of the object will be equal to the total energy.

So, K.E = Total energy

= 3.6 J

Also, we know that

K.E = $\frac{1}{2}mv^{2}_{m}$

or, v = $\sqrt{\frac{2K.E}{m}}$

= $\sqrt{2 \times 3.6 J}{0.2 kg}$

= $\sqrt{36}$

= 6 m/s

thus, we can conclude that maximum speed of the mass during its oscillation is 6 m/s.

4 0

3 years ago